Physics 205B Quiz 2, spring semester 2018
Cuesta College, San Luis Obispo, CA
An object 1.0 cm in height is placed 10 cm in front of a f = –15 cm diverging lens. The absolute value of the image height is:
(A) 0.17 cm.
(B) 0.60 cm.
(C) 0.67 cm.
(D) 1.5 cm.
Correct answer (highlight to unhide): (B)
The image location can be determined from the thin lens equation:
(1/do) + (1/di) = (1/f),
(1/di) = (1/f) – (1/do),
di = ((1/f) – (1/do))–1,
di = ((1/(–15 cm)) – (1/(+10 cm))–1 = –6.0 cm,
Then from the linear magnification equation:
m = hi/ho = –di/do,
the image height can be determined, given the object distance do and image distance di:
hi/ho = –di/do,
hi = –ho·di/do,
hi = –(+1.0 cm)·(–6.0 cm)/(+10 cm) = +0.60 cm.
(Response (A) is the absolute value of (1/f) – (1/do); response (C) is the absolute value of do/f; and response (D) is the absolute value of f/do.)
Sections 30882, 30883
Exam code: quiz02P0wR
(A) : 1 student
(B) : 20 students
(C) : 5 students
(D) : 8 students
Success level: 59%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.63
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